Problem: Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$.

Solution: Assume $O$ to be the center of triangle $ABC$, $OT$ cross $BC$ at $M$, link $XM$, $YM$. Let $P$ be the middle point of $BT$ and $Q$ be the middle point of $CT$, so we have $MT=3\sqrt{15}$. Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\frac{11}{16}$. Notice that $\angle XTY=180^{\circ}-A$, so $\cos XYT=-\cos A$, and this gives us $1143-2XY^2=\frac{-11}{8}XT\cdot YT$. Since $TM$ is perpendicular to $BC$, $BXTM$ and $CYTM$ cocycle (respectively), so $\theta_1=\angle ABC=\angle MTX$ and $\theta_2=\angle ACB=\angle YTM$. So $\angle XPM=2\theta_1$, so\[\frac{\frac{XM}{2}}{XP}=\sin \theta_1\], which yields $XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.$ So same we have $YM=XT$. Apply Ptolemy theorem in $BXTM$ we have $16TY=11TX+3\sqrt{15}BX$, and use Pythagoras theorem we have $BX^2+XT^2=16^2$. Same in $YTMC$ and triangle $CYT$ we have $16TX=11TY+3\sqrt{15}CY$ and $CY^2+YT^2=16^2$. Solve this for $XT$ and $TY$ and submit into the equation about $\cos XYT$, we can obtain the result $XY^2=\boxed{717}$.